https://artofproblemsolving.com/community/c7h1966306_2019_putnam_a2
In the triangle
, let 
be the centroid, and let 
be the center of the inscribed circle. Let 
and 
be the angles at the vertices 
and 
, respectively. Suppose that the segment 
is parallel to 
and that 
. Find .Hey -- what makes this problem so interesting is that it refers to two of the three triangle centers that appear in Common Core Geometry -- the centroid and the incenter. So it was a no-brainer for me to choose this problem to feature in today's post.
I was only able to come up with a partial solution to this problem. Here's how I started -- last year, I remember that a coordinate proof was used to solve the Geometry problem, so maybe we can try something similar this year. (Coordinate proofs appear in Lesson 11-1 of the U of Chicago text.)
We begin by placing one of the vertices at the origin. Since we're given Angle B (beta), let's place point B at the origin. Now we could label the other points A(a, 0) and C(b, c) -- but instead, I decided to label them A(3a, 0) and C(3b, 3c). This is to make it easier to find the centroid G.
In Lesson 11-4 of the U of Chicago text, we learn how to find the center of mass. What's not stated in this lesson is that the centroid of a triangle is simply the center of mass of its vertices -- that is, the average of their coordinates. This allows us to find the coordinates of G very easily:
G((3a + 0 + 3b)/3, (0 + 0 + 3c)/3) = (a + b, c)
This explains why I included those factors of three when selecting coordinates.
Now we are given that beta = 2 arctan (1/3). Notice that the tangent (in the trig sense) of an angle is closely related to slope. So arctan (1/3) refers to a line of slope 1/3 through the origin. But notice that arctan (1/3) doesn't equal beta -- instead, this is beta/2. In other words, the line of slope 1/3 through the origin actually bisects Angle B = beta.
Hmm, angle bisectors -- incenter -- angle bisectors -- incenter -- hey, there's something! The incenter of a triangle is the concurrency point of the angle bisectors. In other words, we do know of a point that lies on the line of slope 1/3 through the origin -- namely I, the incenter of the triangle.
We're also given that
(c - 0)/(x - 0) = (1/3)(x - 0)
x = 3c
So there's our x-coordinate, and so we now have our point I(3c, c). Now we must find the slope of the line passing through A and I. This slope will be negative, and so our desired alpha will equal twice the arctangent of the opposite of this slope:
slope
= c/(3c - 3a)
tan(alpha/2) = c/(3a - 3c)
And this is where I got stuck. We need tan(alpha/2) to be a constant in order to solve the problem, but we're stuck with an expression with two variables, a and c.
Notice that a should be equal to c multiplied by some constant. After all, all triangles with two angles equal to our desired alpha and beta must be similar (by AA~, of course), and so a dilation centered at the origin (B) must map any such triangle to any other such triangle. Thus all coordinates related to these triangles must be proportional. But there's no easy way to find out what a/c is -- it's probably some simple integer (or at least rational) value, but I can't find it.
I tried writing down some other equations to find the coordinates of the incenter I -- if I had them, then I could set the x-coordinate equal to 3c and the y-coordinate equal to c and solve them. But these equations end up being very complicated since they involve distance and square roots.
Unfortunately, I had to give up and peek at the answer thread above. It turns out that the value that I was seeking, c/(3a - 3c), is equal to 1. Once we know this, then we can complete the problem:
tan(alpha/2) = 1
alpha/2 = 45
alpha = 90
So the missing Angle A turns out to be just a right angle! But I didn't know how to find it -- if I had just guessed it was 90, I'd be lucky to get even 1 out of 10 on this problem, considering how tough Putnam graders tend to be. (And if my guess wasn't, my score would be zero.)
One thing that might have helped me is a double-angle formula for tangent. I can never remember what the formula is, and thus I wouldn't have had it if I were a Putnam participant. If I had known the formula, we could use it to find the tangent of beta itself (rather than beta/2), which would give me a slope for
tan(beta) = 2 tan(beta/2)/(1 - tan^2 (beta/2))
= 2(1/3) / (1 - 1/3^2)
= (2/3) / (8/9)
= 3/4
We can use this to rewrite our coordinate C(3b, 3c) in terms of c. We must find the x-coordinate of the point that lies on the line of slope 3/4 through the origin:
(3c - 0)/(x - 0) = (3/4)
x = 4c
So we find the coordinates of C(4c, 3c), thus implying that 3b = 4c and b = 4c/3. But finding a value of b doesn't help us find a value for a, which is what we need to eliminate a in the expression for tan(alpha/2) we found earlier.
When I was working earlier, I'd found a complicated expression for the x-coordinate of the incenter I:
(3a - 3sqrt((a - b)^2 + c^2) + 3(b^2 + c^2))/2
which we set equal to 3c (since we found I(3c, c) earlier) and solve. There's an even more complex expression for the y-coordinate of I. But maybe we can simplify this by plugging in b = 4c/3 that we found using double-angle for tangent earlier. Simplifying gives us the following equation:
3a - 3sqrt((a - c/3)^2 + c^2) = c
Notice that one solver in the Putnam thread JohnDoeSmith was able to find a value for what I call a/b (and what he calls "gamma") -- namely, gamma = 1. Thus a = b = 4c/3. If we plug this into the equation, we see that it indeed gives c = c and checks out. But again, I can't depend on JohnDoeSmith to do the work for me -- I want to find it myself by solving the equation above.
But when I tried to solve this equation by isolating the radical, squaring, and so on, suddenly all the a-terms cancel out, and I'm left with c = 0. Plugging in c = 0 does verify that this is a solution, but this isn't the solution we seek (since if c = 0, we'd have a right angle at B, not A).
And in fact, I couldn't figure out how to obtain the solution a = 4c/3 at all! For some reason, algebraic manipulation resulting in that solution disappearing. (Yes, I did square the equation earlier, but squaring typically adds extraneous solution instead of making solutions disappear.)
And so I give up. Even knowing the tangent double-angle formula didn't lead to a solution. When we look at the thread in more detail, we see that JohnDoeSmith's proof uses another fact:
rs = [ABC]
I believe that r = radius of incircle (that is, the apothem) and s = semiperimeter (especially since awesomemath uses Hero's formula earlier in the thread), and [ABC] is the area of the triangle. I've never seen this formula before (but it does remind me of the regular polygon area formula, since that area is also the product of the apothem and semiperimeter).
I notice that many of the solvers describe their solutions as "bashing." One poster uses "length bashing," while another uses something called "Vincent Huang Bashing." JohnDoeSmith refers to his coordinate proof as "coordinate bashing," and a fourth poster had a "mindless bash." I remember the math Benjamin Leis (in discussing 15-75-90 triangles) referring to an angle "chase" to prove that he indeed had a 15-75-90 triangle. If this had been the Art of Problem Solving website, he likely would have said angle "bash" instead.
Another poster in the thread, Jorvis, mentions using Cartesian coordinates (no bashing, though). He mentions that there was "surprisingly little computation" -- but that doesn't match my experience with this at all. But he never gives a proof. It's possible that his proof is similar to JohnDoeSmith's, where he uses an area formula or something similar to find the missing value.
The poster wen1now links to yet another solution:
https://how-did-i-get-here.com/2019-putnam-a2/
This is also a coordinate proof -- but it's actually on the complex plane. This has the benefit of replacing the tangent double-angle formula with complex multiplication (which is easier to remember as it's just FOIL). But then it appears that the author just guesses that A belongs at 4. He justifies this by noting that Ray CA is the reflection image of CB over the mirror CI, but this isn't obvious. (It's possible that complex multiplication was also used in some way to prove this reflection.)
I also tried to solve Problem A1, since it should be the "easiest." But I also failed to get A1:
https://artofproblemsolving.com/community/c7h1966303_2019_putnam_a1
Determine all possible values of
where , , and 
are nonnegative integers.I figured that the trick is to factor the given expression A^3 + B^3 + C^3 - 3ABC. Then there would be something going on with the factors (even, odd, multiples of three, and so on) that would indicate what the factors might be.
Since the problem involves cubes, I began by cubing the expression (A + B + C):
(A + B + C)^3 = A^3 + B^3 + C^3 + 3(AB^2 + AC^2 + BA^2 + BC^2 + CA^2 + CB^2) + 6ABC
Then I started playing around with this, as in changing some of the signs to negative and hoping that terms would cancel.
As I did this, I thought back to something Ian Stewart wrote in his book about symmetry and deriving the Cubic Formula. Notice that the expression given in the problem:
A^3 + B^3 + C^3 - 3ABC
is symmetric in A, B, and C -- each of the cubes is added, and then we subtract triple the product of all three variables. So if we were to switch A with B, or A with C, or B with C, then the expression would remain the same.
This tells us that any possible factorization of this expression must also symmetric. For example, if one factor is (-A + B + C), then the other two must be (A - B + C) and (A + B - C), or otherwise the product wouldn't be symmetric. I tried this plus linear combinations of this and other expressions, but nothing led to a factorization.
In the thread, it was awesomemath who came up with the correct factorization:
A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 + B^2 + C^2 - AB - BC - CA)
As we can see, this factorization is indeed symmetric in A, B, and C. (Obviously, I'd forgotten to consider one linear factor and one irreducible quadratic factor.) Then awesomemath shows that if the first factor is a multiple of three, then so is the second. Thus expression can't be a multiple of three without being a multiple of nine. (At the Dozenal Forum, we'd say that the expression can't be "singly trine" -- it must be at least "doubly trine" or not trine at all.) Meanwhile, he shows that any number that isn't singly trine is possible by considering equal or consecutive integers for A, B, and C.
Even though I wasn't able to solve any of these Putnam problems on my own, I enjoy looking at the problem-solving techniques that were used here. Who would have thought that I'd use the symmetry mentioned in Stewart's book so soon! If only I'd consider the linear-times-quadratic case, I might have found the correct factorization.
To find the coordinates for C and I in our Problem A-2, we used the Slope Formula in an unusual manner -- we are given a point on a line and a slope, and the y-coordinate of a second point on the line, in order to find its x-coordinate. Last week when I subbed at the seventh grade classroom, I spoke with one of the eighth grade math teachers in the lounge. Her class is currently learning about slope, and she gave her students this exact problem -- that is, find the x-coordinate of a point given its y-coordinate, another point, and the slope of the line between them. (Of course, her students had a number as the y-coordinate and not c or 3c, but they didn't have the benefit of having the origin as the other given point).
Anyway, her students struggled with this problem. She worries that the Common Core is pushing too much algebraic abstraction into eighth grade. On the other hand, traditionalists complain that there isn't enough Algebra I in eighth grade. It could be that Common Core 8 is a poor compromise -- the alternative would be to have a true Algebra I class for eighth graders who are ready and a less abstract math course for those who aren't.
(Then again, I looked at my old Algebra I text -- you know, that's the text with Dolciani listed as one of the authors. In Lesson 8-3, nowhere does Dolciani ask students to find one coordinate of a point given its other coordinate, another point, and a slope. Interpret that as you will.)
Last week, I actually mentioned the Putnam to the seventh graders. My hope in both discussing the exam and telling the moon stories, was to inspire them to want to study more math.
On the other hand, I failed to mention the Putnam in yesterday's Algebra II class. Then again, usually I don't just tell the students about the Putnam -- I give them an example of a Putnam problem. I would have been more motivated to mention the exam if the actual problems had been posted on Sunday night as they usually are, instead of Monday at 5 PM (115 minutes after the final bell rang).
If the problems had been posted before school yesterday, it's possible that I would have shown them Problem A-1, since it's indirectly related to factoring polynomials. That's what the Algebra II students were learning at the time. (I'd much rather mention the Putnam to juniors than seventh graders, since the former is closer to the age when students actually take that exam.)
Lesson 7-7 of the U of Chicago text is called "Sufficient Conditions for Parallelograms." (In the modern Third Edition of the text, this appears as Lesson 7-8.)
This is what I wrote last year about this lesson:
Lesson 7-7 of the U of Chicago text is on sufficient conditions for parallelograms. This is how we prove that a figure is a parallelogram. Dr. Franklin Mason would call these the "parallelogram tests," or, as he often abbreviates it, the "pgram tests."
The sufficient conditions for a parallelogram are that if a quadrilateral has opposite sides congruent, or if it has opposite angles congruent, or if its diagonals bisect each other, or if it has one pair of sides both congruent and parallel, then the figure is a parallelogram.
Think about it. Suppose we are given a quadrilateral with opposite sides congruent. It's easy to use SSS to prove that it's a parallelogram. The two pairs of given sides already give us SS, and for the third S, we draw in a diagonal and note that it's congruent to itself. And so we use the traditional SSS proof to prove this theorem.
In general, if we are given that a figure is symmetrical -- because it's an isosceles triangle or a kite or any figure that has its own Symmetry Theorem -- then we just use that theorem. But if we're given things like congruent sides or angles, then these are probably the S or A parts of one of the traditional Congruence Theorems SSS, SAS, or ASA and so we should use those.
Therefore this section is an excellent demonstration of the power of the SSS, SAS, and ASA Congruence Theorems, even in a Common Core class where the focus is on transformations.
Parallelograms don't have reflection symmetry. On the other hand, they do have rotational symmetry, and here is an example of a proof where rotational symmetry comes in handy:
- Prove that if two perpendicular lines intersect at the center of a square, then they divide that square into four congruent regions.
Another easy case is if the two lines are the perpendicular bisectors of the four sides. Each of the four regions has three right angles (one an angle of the square, and two more right angles by the definition of perpendicular) and hence four right angles (since the sum of the angles of a quadrilateral is 360), and hence each is a rectangle. And as each one's length and height is half that of the original square (by the definition of bisector), hence each is a square. So there are four congruent squares, each with a side length half that of the original square.
But the tough case is when the two lines are neither diagonals nor perpendicular bisectors. The four points where the two lines intersect AB, BC, CD, and DA respectively, let's call these points E, F, G, and H respectively. These quadrilaterals -- AEOH, BFOE. CGOF, and DHOG -- are not special quadrilaterals like kites or trapezoids, yet we must prove them congruent.
Here's my claim -- a 90-degree rotation about O maps AEOH to BFOE. The proof -- we should have proved earlier this week (our extra discussion for Section 7-6) that a rotation maps regular polygons to themselves, so ABCD is mapped to BCDA. Clearly O, the center of rotation, is mapped to itself, so it remains to show where points such as E are mapped. Since EG and FH are perpendicular, rotating line EG must give us line FH, so E'is clearly on line FH. Since E is on AB and AB maps to BC, E' must lie somewhere on line BC. And BC and FH intersect at F, so E' is F.
Similarly, F' is G, G' is H, and H' is E.Therefore the rotation maps AEOH to BFOE -- and since there is an isometry mapping AEOH to BFOE, they are congruent. Similarly both are congruent to CGOF and DHOG. QED
This question was inspired by one I saw many years ago on the Theoni Pappas calendar. Yes, the Pappas calendar I fear I won't be reading any more in three weeks.
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